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Math Help - Limit_Problem Ex_RP

  1. #1
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    Limit_Problem Ex_RP

     lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n*(\frac{2n+1}{n+1})^n
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  2. #2
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    Note that:

    \left(\frac{n+1}{2n}\right) \left(\frac{2n+1}{n+1}\right)

    =\frac{2n+1}{2n}=1+\frac{1}{2n}

    \lim_{n\to \infty}\left(\frac{2n+1}{2n}\right)^{n}=\lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}

    Notice the similarity to the famous 'e' limit?.

    \lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{n}=e

    So, what is the limit of \lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}
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  3. #3
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    Smile

    Hint :
    \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?
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  4. #4
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    Quote Originally Posted by Raoh View Post
    Hint :
    \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?

     e^{\frac{1}{2}}

    but, what about :

    \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n
    Last edited by gilyos; December 19th 2009 at 11:33 AM.
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  5. #5
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    Quote Originally Posted by Raoh View Post
    Hint :
    \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=?

    \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}

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  6. #6
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    Quote Originally Posted by Raoh View Post
    \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}

    therefore your limit must be e^{\frac{1}{2}}.
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  7. #7
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    O.k but what about
    \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???
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  8. #8
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    Quote Originally Posted by gilyos View Post
    O.k but what about
    \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0
    i believe
    \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0
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  9. #9
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    <br />
\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n  = 0<br />
    Why?

    Intuition:

    <br />
(\frac{1}{2} + \frac{1}{2n}) <br />
    is always less than 1, and then raise it to any power greater than one, and it will get smaller.
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  10. #10
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    Quote Originally Posted by gilyos View Post
    O.k but what about
    \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???
    Given the discussion in this thread, a bit of thought might have suggested writing \left( \frac{1}{2} + \frac{1}{2n}\right)^n = \left( \frac{1}{2}\right)^n \left( 1 + \frac{1}{n}\right)^n from which the limit is obvious.
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